PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE
Nouvelle série, tome 98(112) (2015), 199–210
DOI: 10.2298/PIM141009007A
ON THE GROWTH AND THE ZEROS OF SOLUTIONS OF
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS
WITH MEROMORPHIC COEFFICIENTS
Maamar Andasmas and Benharrat Belaïdi
Abstract. We investigate the growth of meromorphic solutions of homogeneous and nonhomogeneous higher order linear differential equations
f (k) +
k−1
X
Aj f (j) + A0 f = 0 (k > 2),
j=1
k−1
f (k) +
X
Aj f (j) + A0 f = Ak (k > 2),
j=1
where Aj (z) (j = 0, 1, . . . , k) are meromorphic functions with finite order.
Under some conditions on the coefficients, we show that all meromorphic solutions f 6≡ 0 of the above equations have an infinite order and infinite lower
order. Furthermore, we give some estimates of their hyper-order, exponent
and hyper-exponent of convergence of distinct zeros. We improve the results
due to Kwon; Chen and Yang; Belaïdi; Chen; Shen and Xu.
1. Introduction and statement of results
In this paper, we use the standard notations of Nevanlinna’s value distribution theory [10, 12, 15]. In addition, we use the notations λ(f ) and λ(1/f ) to
denote respectively the exponents of convergence of the zeros and the poles of a
meromorphic function f , ρ(f ) and µ(f ) to denote respectively the order and the
lower order of f . The hyper-order ρ2 (f ), the hyper-exponent λ2 (f ) of convergence
of zeros and the hyper-exponent λ2 (f ) of convergence of distinct zeros of f are
defined respectively by
log log T (r, f )
log log N (r, 1/f )
ρ2 (f ) = lim sup
, λ2 (f ) = lim sup
,
log
r
log r
r→+∞
r→+∞
2010 Mathematics Subject Classification: Primary 34M10; Secondary 30D35.
Key words and phrases: linear differential equations; meromorphic functions; order of growth;
hyper-order; exponent of convergence of zeros; hyper-exponent of convergence of zeros.
Partially supported by University of Mostaganem (UMAB) (CNEPRU Project Code
B02220120024).
Communicated by Gradimir Milovanović.
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ANDASMAS AND BELAÏDI
log log N (r, 1/f )
.
log r
r→+∞
First, we recall the following definitions. The linear measure of a set E ⊂ (0, +∞) is
R +∞
defined as m(E) = 0 χE (t) dt and the logarithmic measure of a set F ⊂ (1, +∞)
R +∞ 1
is defined by lm(F ) = 1
t χF (t) dt, where χH (t) is the characteristic function of
a set H. The upper density of a set E ⊂ (0, +∞) is defined by
λ2 (f ) = lim sup
dens E = lim sup
r→+∞
m(E ∩ [0, r])
.
r
The upper logarithmic density of a set F ⊂ (1, +∞) is defined by
log dens(F ) = lim sup
r→+∞
lm(F ∩ [1, r])
.
log r
Proposition 1.1. For all H ⊂ [1, +∞) the following statements hold:
i) If lm(H) = ∞, then m(H) = ∞;
ii) If dens H > 0, then m(H) = ∞;
iii) If log dens H > 0, then lm(H) = ∞.
Proof. i) Since we have χH (t)/t 6 χH (t) for all t ∈ H ⊂ [1, +∞), then
m(H) > lm(H).
We can easily prove the results ii) and iii) by applying the definition of the
limit and the properties m(H ∩ [0, r]) 6 m(H) and lm(H ∩ [1, r]) 6 lm(H).
In [11], Kwon investigated the growth of second order equations and obtained
the following result.
Theorem 1.1. [11] Let H be a set of complex numbers satisfying dens{|z| :
z ∈ H} > 0, and let A(z) and B(z) be entire functions such that for real constants
α (> 0), β (> 0),
|A(z)| 6 exp{o(1)|z|β } and |B(z)| > exp{(1 + o(1))α|z|β }
as z → ∞ for z ∈ H. Then every solution f 6≡ 0 of the equation
(1.1)
f ′′ + A(z)f ′ + B(z)f = 0
has infinite order and ρ2 (f ) > β.
In [6], Chen and Yang have studied the growth of solutions of (1.1) and obtained
the following result.
Theorem 1.2. [6] Let H be a set of complex numbers satisfying dens{|z| : z ∈
H} > 0, and let A(z) and B(z) be entire functions with ρ(A) 6 ρ(B) = ρ < +∞
such that for real constant C (> 0) and for any given ε > 0,
|A(z)| 6 exp{o(1)|z|ρ−ε } and |B(z)| > exp{(1 + o(1))C|z|ρ−ε }
as z → ∞ for z ∈ H. Then every solution f 6≡ 0 of equation (1.1) has infinite
order and ρ2 (f ) = ρ(B).
ON THE GROWTH AND THE ZEROS OF SOLUTIONS
201
These results were improved by Belaïdi in [1, 2] by considering more general
conditions to higher order linear differential equations with entire coefficients. Recently in [8], Chen extended the previous results by studying the zeros and the
growth of meromorphic solutions of equation (1.1) and the non-homogeneous equation f ′′ + A(z)f ′ + B(z)f = F when A(z), B(z), F (z) are meromorphic functions.
Here we consider for k > 2 the homogeneous and the non-homogeneous linear
differential equations
(1.2)
f (k) +
k−1
X
Aj f (j) + A0 f = 0,
j=1
(1.3)
f (k) +
k−1
X
Aj f (j) + A0 f = Ak ,
j=1
where Aj (z) (j = 0, 1, . . . , k) (A0 6≡ 0 and Ak 6≡ 0) are meromorphic functions
with finite order. We investigate the zeros and growth of meromorphic solutions of
equations (1.2) and (1.3). The present article may be understood as an extension
and improvement of Theorems 2.3 and 2.4 in the paper of Shen and Xu [14]. We
improve the results due to Chen; Shen and Xu greatly and we give two corollaries
in the case when ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < ρ(A0 ) = σ < 1/2.
Theorem 1.3. Let H ⊂ [0, +∞) be a set with infinite linear measure, and let
Aj (z) (j = 0, 1, . . . , k − 1) be meromorphic functions with finite order. If there exist
positive constants σ > 0, α > 0 such that ρ = max{ρ(Aj ) : j = 1, 2, . . . , k − 1} < σ
σ
and |A0 (z)| > eαr as |z| = r ∈ H, r → +∞, then every meromorphic solution
f 6≡ 0 of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and ρ2 (f ) > σ. Furthermore, if
λ(1/f ) < ∞, then σ 6 ρ2 (f ) 6 ρ(A0 ).
Remark 1.1. By using Proposition 1.1, we can obtain the same results in
Theorem 1.3, while putting H ⊂ [0, +∞) to be a set with positive upper density
(or while putting H ⊂ [1, +∞) to be a set with positive upper logarithmic density)
(or while putting H ⊂ [1, +∞) to be a set with infinite logarithmic measure) instead
of putting H to be a set with infinite linear measure.
Theorem 1.4. Let H ⊂ [0, +∞) be a set with a positive upper density, and
let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions with finite order. If there exist positive constants σ > 0, α > 0 such that ρ = max{ρ(Aj ) :
σ
j = 1, 2, . . . , k} < σ and |A0 (z)| > eαr as |z| = r ∈ H, r → +∞, then every
meromorphic solution f with λ(1/f ) < σ of equation (1.3) is of infinite order and
λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f ).
Furthermore, if λ(1/f ) < min{µ(f ), σ}, then ρ2 (f ) 6 ρ(A0 ).
Remark 1.2. It is clear that ρ(A0 ) = β > σ in Theorems 1.3 and 1.4. Indeed,
suppose that ρ(A0 ) = β < σ. Then, by using Lemma 2.2 of this paper, there exists
a set E1 ⊂ (1, +∞) that has a finite linear measure such that when |z| = r ∈
/
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ANDASMAS AND BELAÏDI
[0, 1] ∪ E1 , r → +∞, we have for any given ε (0 < ε < σ − β)
(1.4)
|A0 (z)| 6 er
β+ε
.
On the other hand, by the hypotheses of Theorems 1.3 and 1.4, there exist positive
constants σ > 0, α > 0 such that
(1.5)
|A0 (z)| > eαr
σ
as |z| = r ∈ H, r → +∞, where H is a set with m(H) = ∞. From (1.4) and (1.5),
we obtain for |z| = r ∈ H r [0, 1] ∪ E1 , r → +∞
σ
eαr 6 |A0 (z)| 6 er
β+ε
and by ε (0 < ε < σ − β) this is a contradiction as r → +∞. Hence ρ(A0 ) = β > σ.
Corollary 1.1. Let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions having only finitely many poles such that ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} <
ρ(A0 ) = σ < 1/2. Then every meromorphic solution f 6≡ 0 of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and ρ2 (f ) = σ. Furthermore, every meromorphic solution
f of equation (1.3) satisfies
λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f )
and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ).
Corollary 1.2. Let Aj (z) (j = 0, 1, . . . , k) (Ak 6≡ 0) be meromorphic functions such that λ(1/A0 ) < µ(A0 ) 6 ρ(A0 ) = σ < 1/2 and ρ = max{ρ(Aj ) : j =
1, 2, . . . , k} < σ. Then every meromorphic solution f 6≡ 0 of equation (1.2) whose
all poles are of uniformly bounded multiplicity satisfies µ(f ) = ρ(f ) = ∞ and
ρ2 (f ) = σ. Furthermore, every meromorphic solution f of equation (1.3) whose all
poles are of uniformly bounded multiplicity satisfies
λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f )
and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ).
2. Auxiliary lemmas
Lemma 2.1. [9] Let f (z) be a transcendental meromorphic function, and let
α > 1, ε > 0 be given constants. Then there exists a set E0 ⊂ [0, ∞) that has
finite linear measure and there exists a constant c > 0, such that for all z satisfying
|z| = r ∈
/ E0 , we have
f (j) (z) 6 c[T (αr, f )rε log T (αr, f )]j (j ∈ N).
f (z)
Lemma 2.2. [5] Let f (z) be a meromorphic function of order ρ(f ) = ρ <
+∞. Then for any given ε > 0, there exists a set E1 ⊂ (1, +∞) that has finite
linear measure and finite logarithmic measure such that when |z| = r ∈
/ [0, 1] ∪ E1 ,
r → +∞, we have |f (z)| 6 exp{rρ+ε }.
P∞
Let g(z) = n=0 an z n be an entire function. We define by µ(r) = max{|an |rn ;
n = 0, 1, · · · } the maximum term of g, and define by νg (r) = max{m; µ(r) =
|am |rm } the central index of g.
ON THE GROWTH AND THE ZEROS OF SOLUTIONS
203
Lemma 2.3. [7] Let f (z) = g(z)/d(z) be a meromorphic function, where g(z)
and d(z) are entire functions such that
µ(g) = µ(f ) = µ 6 ρ(g) = ρ(f ) 6 +∞,
λ(d) = ρ(d) = λ(1/f ) = β < µ.
Let z be a point with |z| = r at which |g(z)| = M (r, g) and νg (r) denote the central
index of g(z). Then there exists a set E2 ⊂ (1, +∞) with finite logarithmic measure
lm(E2 ) < ∞, such that for all z satisfying |z| = r ∈
/ [0, 1] ∪ E2 , we have
f (n) (z) νg (r) n
=
(1 + o(1)) (n > 1 is an integer).
f (z)
z
Lemma 2.4. [6] Let g(z) be an entire function of infinite order, with the hyperorder ρ2 (g) = σ. Then
log log νg (r)
lim sup
= σ.
log r
r→+∞
Lemma 2.5. [13] Let g(z) be an entire function of infinite order. Denote
M (r, g) = max{|g(z)| : |z| = r}, then for any sufficiently large number λ > 0,
and any r ∈ H0 ⊂ (1, +∞)
M (r, g) > c1 exp{c2 rλ },
where lm(H0 ) = ∞ and c1 , c2 are positive constants.
Lemma 2.6. Suppose that k > 2 and h0 , h1 , . . . , hk (h0 6≡ 0) are meromorphic
functions. Let ρ = max{ρ(hj ) : j = 0, 1, . . . , k} < ∞ and let f be a meromorphic
solution of infinite order of the equation
(2.1)
f (k) + hk−1 f (k−1) + · · · + h0 f = hk
with λ(1/f ) = µ < µ(f ). Then ρ2 (f ) 6 ρ.
Proof. We assume that f is a meromorphic solution of infinite order ρ(f ) = ∞
of equation (2.1). We can rewrite (2.1) as
(s) (k) k−1
X
f hk f + .
(2.2)
|hs |
f 6 |h0 | +
f f
s=1
By Hadamard factorization theorem, we can write f as f (z) =
and d(z) are entire functions such that
g(z)
d(z) ,
where g(z)
µ(f ) = µ(g) 6 ρ(f ) = ρ(g) = ∞, λ(d) = ρ(d) = λ(1/f ) = µ < µ(f )
ρ2 (f ) = ρ2 (g).
By Lemma 2.3, there exists a set E2 ⊂ (1, +∞) with finite logarithmic measure,
such that for all z satisfying |z| = r ∈
/ [0, 1] ∪ E2 at which |g(z)| = M (r, g), we have
(n)
f (z)
νg (r) n
(2.3)
=
(1 + o(1)) (n > 1).
f (z)
z
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ANDASMAS AND BELAÏDI
Since ρ(hk (z)d(z)) 6 ρ1 = max{µ, ρ}, then by Lemma 2.2, for any ε > 0 there
exists a set E1 ⊂ (1, +∞) with a finite logarithmic measure such that
(2.4)
|hk (z)d(z)| 6 er
ρ1 +ε
holds for all z satisfying |z| = r ∈
/ [0, 1] ∪ E1 , r → +∞. On the other hand, by
Lemma 2.5 for a sufficiently large number λ > ρ1 , there exist a set H0 ⊂ (1, +∞)
with lm(H0 ) = ∞ and positive constants c1 , c2 such that for any r ∈ H0
(2.5)
M (r, g) > c1 exp{c2 rλ }.
By (2.4) and (2.5), for any given ε with 0 < ε < λ − ρ1 and for all z satisfying
|z| = r ∈ H0 r ([0, 1] ∪ E1 ), r → +∞ at which |g(z)| = M (r, g), we have
ρ1 +ε
h (z) d(z)h (z) er
k k
→ 0, r → +∞.
(2.6)
=
6
f (z)
g(z)
c1 e c 2 r λ
Since ρ = max{ρ (hj ) : j = 0, 1, . . . , k} < ∞, then by Lemma 2.2, we have
(2.7)
|hj (z)| 6 er
ρ+ε
(j = 0, 1, . . . , k)
holds for all z satisfying |z| = r ∈
/ [0, 1] ∪ E1 , r → +∞. Substituting (2.3), (2.6)
and (2.7) into (2.2), we obtain for all z satisfying |z| = r ∈ H0 r ([0, 1] ∪ E1 ∪ E2 ),
r → +∞ at which |g(z)| = M (r, g),
k−1
ν (r) k
X ρ+ε νg (r) s
g r ρ+ε
+
er |1 + o(1)| 6 e
|1 + o(1)| + o(1).
z
z
s=1
So, we get
(2.8)
|νg (r)|k |1 + o(1)| 6 (k + 1)rk er
ρ+ε
|νg (r)|k−1 |1 + o(1)|.
Then, by Lemma 2.4, we obtain from (2.8) that ρ2 (g) = ρ2 (f ) 6 ρ + ε. Since ε
(0 < ε < λ − ρ1 ) being arbitrary, then we get ρ2 (f ) 6 ρ.
Lemma 2.7. [3] Let f (z) be an entire function of order ρ, where 0 < ρ(f ) =
ρ < 1/2, and let ε > 0 be a given constant. Then there exists a set H1 ⊂ [0, +∞)
with dens H1 > 1−2ρ such that |f (z)| > exp{rρ−ε } for all z satisfying |z| = r ∈ H1 .
Lemma 2.8. [7] Suppose that h(z) is a meromorphic function with λ(1/h) <
µ(h) 6 ρ(h) = σ < 1/2. Then for any ε > 0, there exists a set H2 ⊂ (1, +∞) that
has a positive upper logarithmic density such that for all z satisfying |z| = r ∈ H2 ,
we have |h(z)| > exp{(1 + o(1))rσ−ε }.
Lemma 2.9. [11] Let g(z) be a nonconstant entire function of finite order.
Then for any given ε > 0, there exists a set H3 ⊂ [0, +∞) with dens H3 = 1 such
that M (r, g) > exp{rρ(g)−ε } for all z satisfying |z| = r ∈ H3 .
Lemma 2.10. [4] Let Aj (j = 0, 1, . . . , k−1), F 6≡ 0 be finite order meromorphic
functions. If f (z) is an infinite order meromorphic solution of the equation
f (k) + Ak−1 f (k−1) + · · · + A1 f ′ + A0 f = F,
then f satisfies λ(f ) = λ(f ) = ρ(f ) = ∞.
ON THE GROWTH AND THE ZEROS OF SOLUTIONS
205
Lemma 2.11. Let H ⊂ [0, +∞) be a set with a positive upper density (or of
infinite linear measure), and let hj (z) (j = 0, 1, . . . , k) (h0 6≡ 0) be meromorphic
functions with finite order. If there exist positive constants σ > 0, α > 0 such that
σ
|h0 (z)| > eαr as |z| = r ∈ H, r → +∞, and ρ = max{ρ(hj ) : j = 1, 2, . . . , k} < σ,
then every meromorphic solution f 6≡ 0 of equation
(2.9)
f (k) +
k−1
X
hj f (j) + h0 f = hk (k > 2),
j=1
is transcendental and satisfies ρ(f ) > σ.
Proof. Assume that f 6≡ 0 is a meromorphic solution of (2.9) with ρ(f ) < σ.
It follows from (2.9) that
k−1
(2.10)
f (k) X f (j)
hk
−
−
hj
= h0 .
f
f
f
j=1
Since ρ(hj ) < σ (j = 1, 2, . . . , k) and ρ(f ) < σ, then from (2.10) we obtain that the
order of growth of h0 is ρ1 = ρ(h0 ) 6 max{ρ, ρ(f )} < σ. By Lemma 2.2, for any ε
(0 < ε < σ − ρ1 ) there exists a set E1 ⊂ (1, +∞) with a finite linear measure such
that
(2.11)
|h0 (z)| 6 er
ρ1 +ε
holds for all z satisfying |z| = r ∈
/ [0, 1] ∪ E1 , r → +∞. From the hypotheses of
Lemma 2.11, there exists a set H with dens H > 0 (or m(H) = ∞), and there exist
positive constants σ > 0, α > 0 such that
(2.12)
|h0 (z)| > eαr
σ
holds for all z satisfying |z| = r ∈ H, r → +∞. By (2.11) and (2.12), we conclude
σ
ρ1 +ε
that for all z satisfying |z| = r ∈ H r ([0, 1] ∪ E1 ), r → +∞, we have eαr 6 er
and by ε (0 < ε < σ − ρ1 ) this is a contradiction as r → +∞. Consequently,
any meromorphic solution f 6≡ 0 of equation (2.9) is transcendental and satisfies
ρ(f ) > σ.
3. Proofs of the results
Proof of Theorem 1.3. Let f 6≡ 0 be a meromorphic solution of (1.2). It
follows from (1.2) that
(k) k−1
(j) f X
f .
(3.1)
|A0 | 6 +
|Aj |
f
f
j=1
By Lemma 2.11, we know that f is transcendental. By using Lemma 2.1, there
is a set E0 ⊂ (0, +∞) having finite linear measure such that for all z satisfying
|z| = r ∈
/ E0 , we have
(j) f (z) 2k
(3.2)
f (z) 6 r[T (2r, f )] (j = 1, 2, . . . , k).
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ANDASMAS AND BELAÏDI
By Lemma 2.2, for any given ε (0 < ε < σ − ρ) there exists a set E1 ⊂ (1, +∞)
with finite linear measure such that
|Aj (z)| 6 er
(3.3)
ρ+ε
, j = 1, 2, . . . , k − 1
holds for all z satisfying |z| = r ∈
/ [0, 1] ∪ E1 , r → +∞. Also, by the hypotheses of
Theorem 1.3, there exists a set H with m(H) = ∞, such that for all z satisfying
|z| = r ∈ H, r → +∞, we have
σ
|A0 (z)| > eαr .
(3.4)
Hence it follows from (3.1), (3.2), (3.3) and (3.4) that for all z satisfying |z| = r ∈
H r ([0, 1] ∪ E0 ∪ E1 ), r → +∞, we have
(3.5)
σ
eαr 6 r[T (2r, f )]2k +
k−1
X
er
ρ+ε
r[T (2r, f )]2k 6 krer
ρ+ε
[T (2r, f )]2k .
j=1
By 0 < ε < σ − ρ, it follows from (3.5) that
(3.6)
µ(f ) = ρ(f ) = ∞ and ρ2 (f ) > σ.
Furthermore, if λ(1/f ) < ∞, then f is a meromorphic solution of (1.2) with
ρ(f ) = µ(f ) = ∞, λ(1/f ) < µ(f ) and by Remark 1.2, we have max{ρ(Aj ) :
j = 0, 1, . . . , k − 1} = ρ(A0 ) = β < ∞. Thus, by Lemma 2.6, we get
(3.7)
ρ2 (f ) 6 ρ(A0 ).
By (3.6) and (3.7), we conclude that σ 6 ρ2 (f ) 6 ρ(A0 ).
Proof of Theorem 1.4. Let f be a meromorphic solution of (1.3). Assume
that ρ(f ) < ∞. It follows from (1.3) that
(j) (k) k−1
f Ak f X
+
+ .
(3.8)
|A0 | 6 |A
|
j f
f f j=1
By Lemma 2.11, we know that f is transcendental with ρ(f ) > σ. By the hypothesis
λ(1/f ) < σ and Hadamard factorization theorem, we can write f as f (z) = g(z)
d(z) ,
where g(z) and d(z) are entire functions with λ(d) = ρ(d) = λ(1/f ) < σ, ρ(f ) =
ρ(g) > σ. By Lemma 2.9, for any given ε (0 < ε < ρ(f )), there exists a set
H3 ⊂ [0, +∞) with dens H3 = 1 such that
(3.9)
M (r, g) > er
ρ(g)−ε
holds for all z satisfying |z| = r ∈ H3 . By (3.9), for all z satisfying |z| ∈ H3 at
which |g(z)| = M (r, g), we get
(3.10)
|g(z)| > 1.
Then, by (3.10), we obtain
A (z) d(z)A (z) k k
=
6 |d(z)Ak (z)|.
f (z)
g(z)
ON THE GROWTH AND THE ZEROS OF SOLUTIONS
207
We have ρ(d(z)Ak (z)) < σ and set
ρ1 = max{ρ(Aj )(j = 1, 2, . . . , k), ρ(d)} < σ,
hence by using similar arguments as in the proof of Theorem 1.3, for any given ε
(0 < ε < σ − ρ1 ) there exists a set H4 = H ∩ H3 ⊂ [0, +∞) with positive upper
density such that for all z satisfying |z| = r ∈ H4 r ([0, 1] ∪ E0 ∪ E1 ), r → +∞, at
which |g(z)| = M (r, g), we have
(3.13)
f (j) (z) 6 r[T (2r, f )]2k (j = 1, 2, . . . , k),
f (z)
A (z) ρ1 +ε
k ,
6 er
f (z)
(3.14)
|A0 (z)| > eαr .
(3.11)
(3.12)
|Aj (z)| 6 er
ρ1 +ε
(j = 1, 2, . . . , k − 1),
σ
Substituting (3.11)–(3.14) into (3.8), we obtain for all z satisfying |z| = r ∈ H4 r
([0, 1] ∪ E0 ∪ E1 ), r → +∞, at which |g(z)| = M (r, g), that
(3.15)
e
αr σ
2k
6 r[T (2r, f )]
+
k−1
X
r[T (2r, f )]2k + er
ρ1 +ε
6 (k + 1)r[T (2r, f )]2k er
ρ1 +ε
er
ρ1 +ε
j=1
.
Hence by (3.15), we have ρ(f ) = ∞. This is a contradiction which means that the
assumption of ρ(f ) < ∞ is not true. Hence, we conclude that ρ(f ) = ∞.
Since Ak 6≡ 0, then by Lemma 2.10, we obtain λ(f ) = λ(f ) = ρ(f ) = ∞.
By (1.3), it is easy to see that if f has a zero at z0 of order m, m > k, and Aj
(j = 0, 1, . . . , k − 1) are analytic at z0 , then Ak must have a zero at z0 of order
m − k. Therefore, we get by Ak 6≡ 0 that
1
1
1 k−1
X
N r,
6 kN r,
+ N r,
+
N (r, Aj ).
f
f
Ak
j=0
(3.16)
On the other hand, (1.3) may be rewritten as
k−1
1
1 f (k) X f (j)
Aj
=
+
+ A0 .
f
Ak f
f
j=1
So, we get
(3.17)
k
1 k−1
X
X
1
f (j)
6 m r,
+
m(r, Aj ) +
m r,
+ O(1).
m r,
f
Ak
f
j=0
j=1
208
ANDASMAS AND BELAÏDI
By (3.16) and (3.17), we have
(3.18)
1 k−1
1
1
X
6 kN r,
+ N r,
+
N (r, Aj )
T r,
f
f
Ak
j=0
k−1
k
X
1 X
f (j)
+ m r,
+
m(r, Aj ) +
m r,
+ O(1).
Ak
f
j=0
j=1
By (3.18) and the lemma of logarithmic derivative [10], there exists a set E of r of
a finite linear measure such that for all r ∈
/ E, we have
(3.19)
k
1
1 X
T (r, f ) = T r,
+ O(1) 6 kN r,
+
T (r, Aj ) + C log(rT (r, f )),
f
f
j=0
where C is a positive constant. For sufficiently large r, we have
(3.20)
(3.21)
1
T (r, f ).
2
(j = 0, 1, . . . , k),
C log(rT (r, f )) 6
T (r, Aj ) 6 rα+ε
where α = max{ρ(Aj ) (j = 0, 1, . . . , k)}. Then, for r ∈
/ E sufficiently large, by
using (3.20) and (3.21), we conclude from (3.19) that for r ∈
/ E and r sufficiently
large
1
1
+ (k + 1)rα+ε + T (r, f ).
T (r, f ) 6 kN r,
f
2
So, we get
1
(3.22)
T (r, f ) 6 2kN r,
+ 2(k + 1)rα+ε , r ∈
/ E.
f
Hence, by (3.22), we get ρ2 (f ) 6 λ2 (f ) and then λ2 (f ) > λ2 (f ) > ρ2 (f ). Since by
definition, we have λ2 (f ) 6 λ2 (f ) 6 ρ2 (f ), therefore
(3.23)
λ2 (f ) = λ2 (f ) = ρ2 (f ).
Furthermore, if λ(1/f ) < min{µ(f ), σ}, then f is a meromorphic solution of (1.3)
with ρ(f ) = ∞, λ(1/f ) < µ(f ) and by Remark 1.2, we have max{ρ(Aj ) : j =
0, 1, . . . , k} = ρ(A0 ) = β < ∞. Thus, by Lemma 2.6, we get
(3.24)
ρ2 (f ) 6 β.
By (3.23) and (3.24), we conclude that λ2 (f ) = λ2 (f ) = ρ2 (f ) 6 ρ(A0 ).
Proof of Corollary 1.1. Since A0 is a meromorphic function with finitely
many poles and ρ(A0 ) = σ, then by the Hadamard factorization theorem, we can
write A0 as A0 (z) = B(z)
P (z) , where B(z) is an entire function with ρ(A0 ) = ρ(B) = σ
and P (z) is a polynomial. Hence, by Lemma 2.7, for any given ε (0 < ε < σ), there
exists a set H1 ⊂ [0, +∞) with dens H1 > 1 − 2σ > 0 (by Proposition 1.1 we have
ON THE GROWTH AND THE ZEROS OF SOLUTIONS
m(H1 ) = ∞) such that |B(z)| > er
Hence
σ−ε
209
holds for all z, |z| = r ∈ H1 and r → +∞.
σ−ε
(3.25)
|A0 (z)| = |
σ−2ε
B(z)
er
|>
> er
,
m
P (z)
cr
where c > 0 is a constant and m = deg P (z) > 1. Since Aj (j = 0, 1, . . . , k) are
meromorphic functions having finitely many poles and since the poles of f can only
occur at the poles of Aj (j = 0, 1, . . . , k), then f must have finitely many poles.
Therefore, λ(1/f ) = 0. Then, for any given ε with 0 < 2ε < σ − ρ, we have (3.25)
and
(3.26)
ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ − 2ε.
Furthermore, all meromorphic solutions f 6≡ 0 of (1.2) or (1.3) satisfy
(3.27)
λ(1/f ) < σ − 2ε.
Then, from (3.25), (3.26) and (3.27) by application of Theorem 1.3 for equation
(1.2), we find that every meromorphic solution f 6≡ 0 of (1.2) satisfies µ(f ) =
ρ(f ) = ∞ and σ − 2ε 6 ρ2 (f ) 6 σ. By ε (0 < 2ε < σ − ρ) being arbitrary,
we obtain ρ2 (f ) = σ. Furthermore, we conclude from Theorem 1.4 that every
meromorphic solution f of equation (1.3) satisfies λ(f ) = λ(f ) = ρ(f ) = ∞ and
λ2 (f ) = λ2 (f ) = ρ2 (f ) and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ).
Proof of Corollary 1.2. Since A0 is a meromorphic function with λ( A10 ) <
µ(A0 ) 6 ρ(A0 ) = σ < 12 , then by Lemma 2.8, for any ε (0 < ε < σ), there exists a set H2 ⊂ (1, +∞) that has log densH2 > 0 (by Proposition 1.1 we have
m(H2 ) = ∞) such that for all z, |z| = r ∈ H2 , we have
(3.28)
|A0 (z)| > e(1+o(1))r
σ−ε
.
Obviously, the poles of f must occur at the poles of Aj (j = 0, 1, . . . , k), note that
all poles of f are of uniformly bounded multiplicity, then by λ(1/A0 ) < σ and
ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ, we have λ( f1 ) < σ. Then, for a given ε with
0 < ε < min{σ − ρ, σ − λ( f1 )}, we have (3.28) and
(3.29)
ρ = max{ρ(Aj ) : j = 1, 2, . . . , k} < σ − ε and λ(1/f ) < σ − ε.
Then by (3.28) and (3.29) with application of Theorem 1.3 for equation (1.2), we
find that every meromorphic solution f 6≡ 0 whose poles are of uniformly bounded
multiplicity of equation (1.2) satisfies µ(f ) = ρ(f ) = ∞ and σ − ε 6 ρ2 (f ) 6 σ.
Since ε (0 < ε < min{σ − ρ, σ − λ(1/f )}) being arbitrary, we obtain ρ2 (f ) = σ.
Furthermore, with application of Theorem 1.4 for equation (1.3), we find that
every meromorphic solution f whose poles are of uniformly bounded multiplicity
of equation (1.3) satisfies
λ(f ) = λ(f ) = ρ(f ) = ∞, λ2 (f ) = λ2 (f ) = ρ2 (f )
and if λ(1/f ) < µ(f ), then ρ2 (f ) 6 ρ(A0 ).
210
ANDASMAS AND BELAÏDI
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Department of Mathematics
Laboratory of Pure and Applied Mathematics
University of Mostaganem (UMAB)
B. P. 227 Mostaganem
Algeria
maamaths73@yahoo.fr
belaidi@univ-mosta.dz
(Received 22 03 2013)